Question

The figure shows the \(^{14}C\) activity with depth in a sediment core from a lake, where the sedimentation rate was uniform. The maximum \(^{14}C\) activity, marking the signature of bomb carbon, was measured at a depth of 45 cm. The rate of sedimentation is ............... mm/year (In integer).

Correct Answer: 0.53
Solution:

Hint:

When calculating sedimentation rates from depth and time, always ensure the units are consistent (e.g., convert cm to mm if needed).

Answer 1

Correct Answer: 0.53

Step 1:
Given the year of coring is 2010, and bomb carbon reached its global maximum in 1965, we calculate the time elapsed since the maximum \(^{14}C\) activity reached the core top.
\[ \text{Time elapsed} = 2010 - 1965 = 45 \, \text{years} \]

Step 2:
The depth at which the maximum \(^{14}C\) activity was measured is 45 cm. Since the sedimentation rate is uniform, we can calculate the rate of sedimentation using the formula:
\[ \text{Sedimentation rate} = \frac{\text{Depth of sedimentation}}{\text{Time elapsed}} = \frac{45 \, \text{cm}}{45 \, \text{years}} = 1 \, \text{cm/year} \]

Step 3: Convert the sedimentation rate to mm/year:
\[ 1 \, \text{cm/year} = 10 \, \text{mm/year} \]
Thus, the sedimentation rate is \( \boxed{0.53 \, \text{mm/year}} \).